Aniline has Kb = 4.3 x 10^-10. What is the pH of a 0.233 M solution of aniline hydrochloride?

1. The salt supplies the anilinium cation, which is the conjugate acid of aniline. 2. Chloride is a spectator, so only the cation affects the pH. 3. Find its acid constant: 1.0 x 10^-14 divided by 4.3 x 10^-10 = 2.3 x 10^-5. 4. Set up the usual weak acid problem with 0.233 M starting concentration. 5. Assume the change is small: x squared = 0.233 x 2.3 x 10^-5 = 5.4 x 10^-6. 6. Take the square root: x = 2.3 x 10^-3 mol per litre of hydronium ion. 7. That is about 1 percent of 0.233, so the assumption holds comfortably. 8. Take the negative logarithm: pH = 2.64, an acidic solution as expected for this salt. _Source: OpenStax Chemistry (CC BY 4.0), Ch 14 "Acid-Base Equilibria", section 14.4 Hydrolysis of Salt Solutions_